Solve for $n$, $ \dfrac{3n - 3}{n - 2} = -\dfrac{4}{4n - 8} + \dfrac{2}{n - 2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n - 2$ $4n - 8$ and $n - 2$ The common denominator is $4n - 8$ To get $4n - 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{3n - 3}{n - 2} \times \dfrac{4}{4} = \dfrac{12n - 12}{4n - 8} $ The denominator of the second term is already $4n - 8$ , so we don't need to change it. To get $4n - 8$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{2}{n - 2} \times \dfrac{4}{4} = \dfrac{8}{4n - 8} $ This give us: $ \dfrac{12n - 12}{4n - 8} = -\dfrac{4}{4n - 8} + \dfrac{8}{4n - 8} $ If we multiply both sides of the equation by $4n - 8$ , we get: $ 12n - 12 = -4 + 8$ $ 12n - 12 = 4$ $ 12n = 16 $ $ n = \dfrac{4}{3}$